30/x-y+4y/x+y=10 40/x-y+55/x+y=13
Dear Student,
I assume your question as below and have provided the solution, as it appears to be faulty.
Given : + = 10
- = 13
To find : the values of x and y
Let be a and be b
then,
=> 44a +30b =10 .................(1)
and 55a +40b=13..................(2)
Multiply eq (1) by 4 and eq (2) by 3
=> 176a +120b =40 .................(3)
and 165a +120b=39..................(4)
Now subtract eq (3) by eq 4 , we get
11a = 1
=> a = 1/ 11
applying this value of a in eq 1 we get
=> 44 + 30 b = 10
=> 30 b = 6
=> b = 1/5
Now as a = and b =
=> x+y = 11............(5)
and x-y = 5.............(6)
adding eq (5) and eq (6)
=>2x = 16
=> x= 8
similarily applying this value in eq (5) we get
=> 8+ y = 11
=> y = 3
Regards.
I assume your question as below and have provided the solution, as it appears to be faulty.
Given : + = 10
- = 13
To find : the values of x and y
Let be a and be b
then,
=> 44a +30b =10 .................(1)
and 55a +40b=13..................(2)
Multiply eq (1) by 4 and eq (2) by 3
=> 176a +120b =40 .................(3)
and 165a +120b=39..................(4)
Now subtract eq (3) by eq 4 , we get
11a = 1
=> a = 1/ 11
applying this value of a in eq 1 we get
=> 44 + 30 b = 10
=> 30 b = 6
=> b = 1/5
Now as a = and b =
=> x+y = 11............(5)
and x-y = 5.............(6)
adding eq (5) and eq (6)
=>2x = 16
=> x= 8
similarily applying this value in eq (5) we get
=> 8+ y = 11
=> y = 3
Regards.